form.ajaxSubmit获取上传的文件名

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    var form = $("form[name=fileForm]");    $("#uploadTip").html("正在上传...");    var options = {        action: '/optimizationJob/uploadFile.action',        type: 'post',        data: {fileName:fileName },//传递文件名到服务器        success: function (data) {            var success = data.success;            var errMsg = data.errMsg;            if (success == "Y") {                console.log("上传成功,返回success=Y,errMsg:" + errMsg);                $("#uploadTip").html("上传成功");                // uploadFileToJSS(fileName);            } else {                console.log("上传失败,返回success=N,errMsg:" + errMsg);                $("#uploadTip").html("上传失败");            }            $("#submitBtn").attr("disabled", false);        },        error: function (data) {            console.log("上传失败,返回error");            $("#uploadTip").html("上传失败");            $("#submitBtn").attr("disabled", false);        }    };    form.ajaxSubmit(options);

java中的Action写好fileName的set,get方法，就能使用了

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